I really don't know why youtube is not recommending your videos in the top list...excellenet way of explaining the problem and the solution...starting with brute force method to optimized and then dry-run of the code...best videos ever...keep the videos rolling....don't get bogged down by the view counts...

A very few can explain with real life examples and I am happy to find this channel which are very informative👏

I don't usually leave comments. . So far, I haven't seen such a thorough explanation. thank you very much, sir

Bro you are truly a Gem for the coder community your explanation and teaching of each point are very understandable form.

I was stuck in this problem almost a day to solve without transposing, after watching this I was clear now, Thanks bro, keep uploading videos

Thank you! You are the best man!

this explanation helped me a lot

much appreciated Sir, thanks.

in the second loop why you didnt use (n+1)/2 instead of n/2 ?

thanks for these ✨

you have not explained how you came up with all the indices, however thanks for the final thoughts!

You solution is easy and straight forward. One question, why are you running loops (n+1)/2 and n/2? Thanks

of immense help

This solution is optimal, but it just feels too heavy to list all the indices in a cycle of swappings. So, I provide another solution using only the information that we need to carry each value at index (i, j) to the new index (j, n - 1 - i). It's easy to observe that to do a cycle of swappings, we just need to swap the first element to the second element, then the first element to the third element, ..., and finally the first element to the last element. Then, the code becomes (python3): int n = len(matrix); def nextPoint(x, y): return y, n - 1 - x for i in range((n + 1) // 2): for j in range(n // 2): x, y = nextPoint(i, j) while x != i or y != j: matrix[i][j], matrix[x][y] = matrix[x][y], matrix[i][j] x, y = nextPoint(x, y) This method works with all types of permutation.

Thanks sir😇

As always very descriptive and neat explanation bro👍 Regarding the loop bounds, i think it would be best the inner loop(j) runs for +1 iteration, instead of the outer loop(i). So conceptually, from your logical explanation, for each ring which is the ith loop, all elements in the ring get rotated as part of the j iterations, then we move to the inner ring etc. But if we keep i loop for more iteration(N+1/2), for odd matrixes like 3*3 say, the inner ring is just 1 element, so ideally there shouldn't be a second ring to be rotated/fixed (as its only 1 element matrix[1][1]). but we can see here that the last rotation in the outer loop is happening as part of the inner ring (i=1) iteration (which ideally should have got completed as part of i=0 itself). If we keep j range from 0 to (N+1)/2, this will get fixed. Hope that makes sense.

How u define limits as n+1/2 and n/2 respectively explain pls😢

100% effect 100% Clarity

i understand why n/2 because in the inner layer we only need to do half, but i dont understand why outer layer is n+1/2 , shouldn't they be the same since column and row are the same dims?

Can you please explain how will the elements 9,7,14,2 get swapped when we are running the inner loop only till j < n/2 which in this case would mean that j=0 or j=1. What happens to the elements that are at j=2. How will they get swapped?

I think the time complexity must be O(N) , as each cell is getting read once and written once.

Could someone explain why the final elements (9, 7, 14, 2) in outer ring will get swapped, when we moved to inner ring i.e., i=1 but not when we are iterating at outer ring i.e., i=0 ?

Brother can you explain for me why you know the syntax concept at each position, I understand thr concept you taugh but don't know how to write it in code ☹️☹️ how to know and write syntax at each position i wanna write 😢 so hard for me bro 😭, what is matrix[n-1-j] and matrix[n-1-i][n-j-1], matrix[j][n-1-i], and mattix[i][j] so hard brother

hi Nikhil Lohia, i have a question, how to rotate matrix 45 degrees so the matrix size will increase, i trying to make the solution and now i want to give up (i have tried straight 5 hours btw). Can you make this solution a video pls.

Can you explain the position of indices , its confusing where to consider n-1-i Or n-j-1 . Please explain Otherwise best explanation of the question

how to know the position likr bottom left is matrix[n - 1 - j][i]? It's hard to understand that and why above for loop (i < (n+1) / 2) and the second loop (j < n / 2), how do you know the syntax at each posititon top left top right bottom left bottom right, it's hard to understand 😢 can anybody explain how to know syntax at each posititon, omg so hard 😢☹️

ok simple firstly Transpose the given matrix and then reverse it. ------------------------------------------------------- Runtime 0 ms Beats 100.00% of users with Java

sir you are doing awesome but your voice of video is very low